1. circle the brønsted - lowry acids and underline the brønsted - lowry bases: hcl po₄³⁻ ch₃nh₂…

Explanation:

Step1: Recall Bronsted - Lowry definitions

A Bronsted - Lowry acid is a proton donor and a Bronsted - Lowry base is a proton acceptor.

Step2: Analyze each species

• $HCl$: It can donate a proton ($H^+$), so it is a Bronsted - Lowry acid. Circle $HCl$.
• $PO_{4}^{3 - }$: It can accept a proton to form $HPO_{4}^{2 - }$, so it is a Bronsted - Lowry base. Underline $PO_{4}^{3 - }$.
• $CH_{3}NH_{2}$: It has a lone pair of electrons on the nitrogen atom and can accept a proton to form $CH_{3}NH_{3}^{+}$, so it is a Bronsted - Lowry base. Underline $CH_{3}NH_{2}$.
• $CH_{3}(CH_{2}){2}COOH$: The carboxyl group $-COOH$ can donate a proton, so it is a Bronsted - Lowry acid. Circle $CH{3}(CH_{2})_{2}COOH$.
• $S^{2 - }$: It can accept a proton to form $HS^{-}$, so it is a Bronsted - Lowry base. Underline $S^{2 - }$.
• $NH_{4}^{+}$: It can donate a proton to form $NH_{3}$, so it is a Bronsted - Lowry acid. Circle $NH_{4}^{+}$.

Answer:

Circle: $HCl$, $CH_{3}(CH_{2}){2}COOH$, $NH{4}^{+}$; Underline: $PO_{4}^{3 - }$, $CH_{3}NH_{2}$, $S^{2 - }$